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 March 8th, 2017, 11:02 AM #1 Newbie   Joined: Mar 2017 From: uk Posts: 1 Thanks: 0 Solve trigonometry problem Hello, Please help. I have been given the question: Find all solutions (in degrees) in the range âˆ’360Â° < Î¸ < 360Â° such that cos(4Î¸) = cos(Î¸) I have no clue how to even approach this question, can someone please help me? Thanks in advance. Last edited by skipjack; March 8th, 2017 at 11:22 AM.
 March 8th, 2017, 11:32 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,045 Thanks: 1395 The equation implies that 4Î¸ = Â±Î¸ + k*360Â°, where k is an integer. Hence Î¸ is an integer multiple of 72Â° or 120Â°. List the 13 solutions that lie in the specified interval.
March 8th, 2017, 02:56 PM   #3
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Quote:
 Originally Posted by Heath2017 Hello, Please help. I have been given the question: Find all solutions (in degrees) in the range âˆ’360Â° < Î¸ < 360Â° such that cos(4Î¸) = cos(Î¸) I have no clue how to even approach this question, can someone please help me? Thanks in advance.
\displaystyle \begin{align*} \cos{ \left( 4\,x \right) } &= \cos{ \left( x \right) } \\ 2\, \left[ \cos{ \left( 2\,x \right) } \right] ^2 - 1 &= \cos{ \left( x \right) } \\ 2\,\left[ 2\cos^2{ \left( x \right) } - 1 \right] ^2 - 1 &= \cos{ (x)} \\ 2\,\left[ 4\cos^4{(x)} - 4\cos^2{(x)} + 1 \right] - 1 &= \cos{(x)} \\ 8\cos^4{(x)} - 8\cos^2{(x)} + 2 - 1 &= \cos{(x)} \\ 8\cos^4{(x)} - 8\cos^2{(x)} - \cos{(x)} + 1 &= 0 \\ 8\cos^2{(x)}\,\left[ \cos^2{(x)} - 1 \right] - \cos{(x)} + 1 &= 0 \\ 8\cos^2{(x)}\,\left[ \cos{(x)} - 1 \right] \left[ \cos{(x)} + 1 \right] -1\,\left[ \cos{(x)} - 1 \right] &= 0 \\ \left[ \cos{(x)} - 1 \right] \,\left\{ 8\cos^2{(x)}\,\left[ \cos{(x)} + 1 \right] - 1 \right\} &= 0 \\ \left[ \cos{(x)} - 1 \right] \left[ 8\cos^3{(x)} + 8\cos^2{(x)} - 1 \right] &= 0 \end{align*}

This factors further. Hint: \displaystyle \begin{align*} \cos{(x)} = -\frac{1}{2} \end{align*} is another solution.

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